Normal distribution and confidence intervals

Hello,

I am looking for a little guidance or help with the following question:

*The standard deviation of the masses of 600 blocks is 248 kg. *

*A random sample of 55 blocks has a mean mass of 4.72 Mg.*

*With what degree of confidence can it be said that the mean mass of all the blocks is 4.72 **± 0.100 Mg?*

I've tried reading up some information on the normal distribution table but can't really get my head around it. Could anyone offer me some advice or provide me with a good link with information on the subject?

Thanks.

Re: Normal distribution and confidence intervals

I am not an expert on this stuff, but I think you need to figure out the standard error for your sample. Once you know standard error, you can use it to judge how close your estimate for the mean should be and with what confidence level. Essentially, standard error is the standard deviation of your sample. So, if you were to find the mean mass of every possible sample (from 1 block, all the way up to 600 blocks), you should be able to calculate the approximate standard deviation of the set of samples based on the actual standard deviation of the masses and the size of your sample. So, using the binomial theorem, since your sample is greater than 30 blocks, you can use the formula that standard error is standard deviation over the square root of your sample size.

$\displaystyle \text{St Err}=\frac{\sigma}{\sqrt{n}}$

Where $\displaystyle \sigma$ is your standard deviation of 0.248Mg=248kg, and your sample size $\displaystyle n=55$. Need any more assistance?

Re: Normal distribution and confidence intervals

Quote:

Originally Posted by

**StephenB1965** *The standard deviation of the masses of 600 blocks is 248 kg. *

*A random sample of 55 blocks has a mean mass of 4.72 Mg.*

*With what degree of confidence can it be said that the mean mass of all the blocks is 4.72 **± 0.100 Mg?*...

Assuming a normally distributed population without replacement samples:

Population size = N =600

Sample size =n = 55

$\displaystyle \sigma =248$

then, sample standard error:

$\displaystyle s=\frac{\sigma }{\sqrt{n}}*\sqrt{\frac{\text{N}-n}{\text{N}-1}} = 31.897$

and z value is:

$\displaystyle z=\pm \frac{100}{s}=\pm 3.1351$

Now find probability from a normal distribution table.

Re: Normal distribution and confidence intervals

Many thanks but I thought the sample standard error was calculated by:

http://onlinestatbook.com/chapter8/graphics/sem_pop.gif

Re: Normal distribution and confidence intervals

StephenB1965,

I am also looking for how to figure out confidence intervals when I only have a mean of a sample and a SD of the population. If you get any luck, drop me a message :) will keep checking this post though.

Also, I thought the standard error of the mean was

SE = Standard Error of Population Mean

SD = Standard Deviation of Population

SE = SD/sqrrt(SampleSize)

Re: Normal distribution and confidence intervals

Quote:

Originally Posted by

**MaxJasper** and z value is:

$\displaystyle z=\pm \frac{100}{s}=\pm 3.1351$

Now find probability from a normal distribution table.

MaxJasper, what is the z value for?

Re: Normal distribution and confidence intervals

Quote:

Originally Posted by

**JungleMath** MaxJasper, what is the z value for?

http://www.pearsonclinical.co.uk/Sit...stribution.jpg

Re: Normal distribution and confidence intervals

Thank you very much, helped out alot.

Re: Normal distribution and confidence intervals

Thanks for your help so far MaxJasper, one quick query tho, to get the s value (which I presume is the sample standard error) why do you have to multiply the standand error by √N – n / N - 1?

Re: Normal distribution and confidence intervals

Normal distribution assumes N=infinite, when N<inf then the modified version is used for sample standard error.